Showing that $f(x)=x$ for at least one point in $[0,1]$












0












$begingroup$


Problem: Let $I=[0,1]$ be the closed unit interval. Suppose $f$ is a continuous mapping of $I$ into $I$. Prove that $f(x)=x$ for at least one $x in I$.



Attempt:



We have a known result:



Let $g$ be a continuous real valued function on a metric space $X$. Then $Z(g)={p in X: g(p)=0 }$, i.e. the zero set of $g$ is closed. [Proof is simple using "inverse image of a closed set is closed under continuous map"]



Now, we construct $F:I to I$, such that $F(x) =f(x)-x$, which is definitely continuous, being a linear combination of two continuous functions.

[It is assumed that $F(x)=f(x)-x>0$, $forall xin I$. [ If $x>f(x)$, consider the function $F_1(x)=x-f(x)$ ]. Otherwise, if the function $F(x)$ changes sign somewhere within the interval, it must attain the value $0$, giving us $f(x)=x$ ].



But we know that $Z(F)$ is closed, hence it cannot be $phi$. We are done.



Is this at all a valid proof?



Edit: Being doubtful, I write up another approach:



The function $f$ maps into $I$, i.e. itself. Hence, to be $f(x)>x$ for every value of $x$, its range set would have to exceed $I$, and the best case scenario would be the identity map, for which $f(x)=x$ for all $x$ . Otherwise, $F(x)$ must change sign.






real-analysis proof-verification continuity metric-spaces







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Problem: Let $I=[0,1]$ be the closed unit interval. Suppose $f$ is a continuous mapping of $I$ into $I$. Prove that $f(x)=x$ for at least one $x in I$.



    Attempt:



    We have a known result:



    Let $g$ be a continuous real valued function on a metric space $X$. Then $Z(g)={p in X: g(p)=0 }$, i.e. the zero set of $g$ is closed. [Proof is simple using "inverse image of a closed set is closed under continuous map"]



    Now, we construct $F:I to I$, such that $F(x) =f(x)-x$, which is definitely continuous, being a linear combination of two continuous functions.

    [It is assumed that $F(x)=f(x)-x>0$, $forall xin I$. [ If $x>f(x)$, consider the function $F_1(x)=x-f(x)$ ]. Otherwise, if the function $F(x)$ changes sign somewhere within the interval, it must attain the value $0$, giving us $f(x)=x$ ].



    But we know that $Z(F)$ is closed, hence it cannot be $phi$. We are done.



    Is this at all a valid proof?



    Edit: Being doubtful, I write up another approach:



    The function $f$ maps into $I$, i.e. itself. Hence, to be $f(x)>x$ for every value of $x$, its range set would have to exceed $I$, and the best case scenario would be the identity map, for which $f(x)=x$ for all $x$ . Otherwise, $F(x)$ must change sign.






    real-analysis proof-verification continuity metric-spaces







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      Problem: Let $I=[0,1]$ be the closed unit interval. Suppose $f$ is a continuous mapping of $I$ into $I$. Prove that $f(x)=x$ for at least one $x in I$.



      Attempt:



      We have a known result:



      Let $g$ be a continuous real valued function on a metric space $X$. Then $Z(g)={p in X: g(p)=0 }$, i.e. the zero set of $g$ is closed. [Proof is simple using "inverse image of a closed set is closed under continuous map"]



      Now, we construct $F:I to I$, such that $F(x) =f(x)-x$, which is definitely continuous, being a linear combination of two continuous functions.

      [It is assumed that $F(x)=f(x)-x>0$, $forall xin I$. [ If $x>f(x)$, consider the function $F_1(x)=x-f(x)$ ]. Otherwise, if the function $F(x)$ changes sign somewhere within the interval, it must attain the value $0$, giving us $f(x)=x$ ].



      But we know that $Z(F)$ is closed, hence it cannot be $phi$. We are done.



      Is this at all a valid proof?



      Edit: Being doubtful, I write up another approach:



      The function $f$ maps into $I$, i.e. itself. Hence, to be $f(x)>x$ for every value of $x$, its range set would have to exceed $I$, and the best case scenario would be the identity map, for which $f(x)=x$ for all $x$ . Otherwise, $F(x)$ must change sign.






      real-analysis proof-verification continuity metric-spaces







      share|cite|improve this question











      $endgroup$




      Problem: Let $I=[0,1]$ be the closed unit interval. Suppose $f$ is a continuous mapping of $I$ into $I$. Prove that $f(x)=x$ for at least one $x in I$.



      Attempt:



      We have a known result:



      Let $g$ be a continuous real valued function on a metric space $X$. Then $Z(g)={p in X: g(p)=0 }$, i.e. the zero set of $g$ is closed. [Proof is simple using "inverse image of a closed set is closed under continuous map"]



      Now, we construct $F:I to I$, such that $F(x) =f(x)-x$, which is definitely continuous, being a linear combination of two continuous functions.

      [It is assumed that $F(x)=f(x)-x>0$, $forall xin I$. [ If $x>f(x)$, consider the function $F_1(x)=x-f(x)$ ]. Otherwise, if the function $F(x)$ changes sign somewhere within the interval, it must attain the value $0$, giving us $f(x)=x$ ].



      But we know that $Z(F)$ is closed, hence it cannot be $phi$. We are done.



      Is this at all a valid proof?



      Edit: Being doubtful, I write up another approach:



      The function $f$ maps into $I$, i.e. itself. Hence, to be $f(x)>x$ for every value of $x$, its range set would have to exceed $I$, and the best case scenario would be the identity map, for which $f(x)=x$ for all $x$ . Otherwise, $F(x)$ must change sign.






      real-analysis proof-verification continuity metric-spaces




      real-analysis proof-verification continuity metric-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 16 at 20:44







      Subhasis Biswas

















      asked Jan 16 at 20:34









      Subhasis BiswasSubhasis Biswas

      515412




      515412






















          2 Answers
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          active

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          $begingroup$

          No, it is not at all valid. The empty set is also closed.



          Note that your "proof" would also apply to $F(x) = f(x) - x/2$, but there is not necessarily a solution to $f(x) = x/2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you please check my "proof" now?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:47










          • $begingroup$
            Not only that, using my argument would be so disastrous, that we can "Prove" every continuous function attains zero, which is ridiculous.
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:51





















          3












          $begingroup$

          No, the proof is not valid. It is not clear (nor is it necessarily true) that $F$ is a mapping from $I$ to $I$.



          I would approach it this way: Letting $F$ be as you specified, it follows that $F(0) geq 0$ while $F(1) le 0$. Furthermore, $F$ is also continuous. Therefore by the intermediate value theorem there is an $x in [0,1] = I$ s.t. $F(x) = 0$, or equvialently, $f(x) = x$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            can you check mine now?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:45










          • $begingroup$
            Well, the writing really does need to be improved. And, the inequality $f(x) > x$ could hold for every point in $I setminus {1}$. So I would advise you to try again.
            $endgroup$
            – Mike
            Jan 16 at 20:50










          • $begingroup$
            The domain is compact!? So we cannot leave out the boundary point.
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:53










          • $begingroup$
            Does the result hold good in $(0,1)$? Can you help me out?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:56










          • $begingroup$
            To answer your question, the result does not necessarily hold in $(0,1)$. For example, consider the function $f(x)= 1/2 + x/2$. The idea for showing that it does hold in $[0,1]$ is this: The function $F$ is nonnegative at $x=0$ and it is nonpositive at $x=1$. so as $F$ is also continuous, somewhere in $[0,1]$ (even if it is at precisely one endpoint $x=0$ or the other endpoint $x=1$ or at both endpoints of $I$, it must take on the value 0.
            $endgroup$
            – Mike
            Jan 16 at 20:59














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          2 Answers
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          2 Answers
          2






          active

          oldest

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          active

          oldest

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          active

          oldest

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          3












          $begingroup$

          No, it is not at all valid. The empty set is also closed.



          Note that your "proof" would also apply to $F(x) = f(x) - x/2$, but there is not necessarily a solution to $f(x) = x/2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you please check my "proof" now?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:47










          • $begingroup$
            Not only that, using my argument would be so disastrous, that we can "Prove" every continuous function attains zero, which is ridiculous.
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:51


















          3












          $begingroup$

          No, it is not at all valid. The empty set is also closed.



          Note that your "proof" would also apply to $F(x) = f(x) - x/2$, but there is not necessarily a solution to $f(x) = x/2$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Could you please check my "proof" now?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:47










          • $begingroup$
            Not only that, using my argument would be so disastrous, that we can "Prove" every continuous function attains zero, which is ridiculous.
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:51
















          3












          3








          3





          $begingroup$

          No, it is not at all valid. The empty set is also closed.



          Note that your "proof" would also apply to $F(x) = f(x) - x/2$, but there is not necessarily a solution to $f(x) = x/2$.






          share|cite|improve this answer









          $endgroup$



          No, it is not at all valid. The empty set is also closed.



          Note that your "proof" would also apply to $F(x) = f(x) - x/2$, but there is not necessarily a solution to $f(x) = x/2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 16 at 20:36









          Robert IsraelRobert Israel

          331k23221477




          331k23221477












          • $begingroup$
            Could you please check my "proof" now?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:47










          • $begingroup$
            Not only that, using my argument would be so disastrous, that we can "Prove" every continuous function attains zero, which is ridiculous.
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:51




















          • $begingroup$
            Could you please check my "proof" now?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:47










          • $begingroup$
            Not only that, using my argument would be so disastrous, that we can "Prove" every continuous function attains zero, which is ridiculous.
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:51


















          $begingroup$
          Could you please check my "proof" now?
          $endgroup$
          – Subhasis Biswas
          Jan 16 at 20:47




          $begingroup$
          Could you please check my "proof" now?
          $endgroup$
          – Subhasis Biswas
          Jan 16 at 20:47












          $begingroup$
          Not only that, using my argument would be so disastrous, that we can "Prove" every continuous function attains zero, which is ridiculous.
          $endgroup$
          – Subhasis Biswas
          Jan 16 at 20:51






          $begingroup$
          Not only that, using my argument would be so disastrous, that we can "Prove" every continuous function attains zero, which is ridiculous.
          $endgroup$
          – Subhasis Biswas
          Jan 16 at 20:51













          3












          $begingroup$

          No, the proof is not valid. It is not clear (nor is it necessarily true) that $F$ is a mapping from $I$ to $I$.



          I would approach it this way: Letting $F$ be as you specified, it follows that $F(0) geq 0$ while $F(1) le 0$. Furthermore, $F$ is also continuous. Therefore by the intermediate value theorem there is an $x in [0,1] = I$ s.t. $F(x) = 0$, or equvialently, $f(x) = x$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            can you check mine now?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:45










          • $begingroup$
            Well, the writing really does need to be improved. And, the inequality $f(x) > x$ could hold for every point in $I setminus {1}$. So I would advise you to try again.
            $endgroup$
            – Mike
            Jan 16 at 20:50










          • $begingroup$
            The domain is compact!? So we cannot leave out the boundary point.
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:53










          • $begingroup$
            Does the result hold good in $(0,1)$? Can you help me out?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:56










          • $begingroup$
            To answer your question, the result does not necessarily hold in $(0,1)$. For example, consider the function $f(x)= 1/2 + x/2$. The idea for showing that it does hold in $[0,1]$ is this: The function $F$ is nonnegative at $x=0$ and it is nonpositive at $x=1$. so as $F$ is also continuous, somewhere in $[0,1]$ (even if it is at precisely one endpoint $x=0$ or the other endpoint $x=1$ or at both endpoints of $I$, it must take on the value 0.
            $endgroup$
            – Mike
            Jan 16 at 20:59


















          3












          $begingroup$

          No, the proof is not valid. It is not clear (nor is it necessarily true) that $F$ is a mapping from $I$ to $I$.



          I would approach it this way: Letting $F$ be as you specified, it follows that $F(0) geq 0$ while $F(1) le 0$. Furthermore, $F$ is also continuous. Therefore by the intermediate value theorem there is an $x in [0,1] = I$ s.t. $F(x) = 0$, or equvialently, $f(x) = x$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            can you check mine now?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:45










          • $begingroup$
            Well, the writing really does need to be improved. And, the inequality $f(x) > x$ could hold for every point in $I setminus {1}$. So I would advise you to try again.
            $endgroup$
            – Mike
            Jan 16 at 20:50










          • $begingroup$
            The domain is compact!? So we cannot leave out the boundary point.
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:53










          • $begingroup$
            Does the result hold good in $(0,1)$? Can you help me out?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:56










          • $begingroup$
            To answer your question, the result does not necessarily hold in $(0,1)$. For example, consider the function $f(x)= 1/2 + x/2$. The idea for showing that it does hold in $[0,1]$ is this: The function $F$ is nonnegative at $x=0$ and it is nonpositive at $x=1$. so as $F$ is also continuous, somewhere in $[0,1]$ (even if it is at precisely one endpoint $x=0$ or the other endpoint $x=1$ or at both endpoints of $I$, it must take on the value 0.
            $endgroup$
            – Mike
            Jan 16 at 20:59
















          3












          3








          3





          $begingroup$

          No, the proof is not valid. It is not clear (nor is it necessarily true) that $F$ is a mapping from $I$ to $I$.



          I would approach it this way: Letting $F$ be as you specified, it follows that $F(0) geq 0$ while $F(1) le 0$. Furthermore, $F$ is also continuous. Therefore by the intermediate value theorem there is an $x in [0,1] = I$ s.t. $F(x) = 0$, or equvialently, $f(x) = x$.






          share|cite|improve this answer









          $endgroup$



          No, the proof is not valid. It is not clear (nor is it necessarily true) that $F$ is a mapping from $I$ to $I$.



          I would approach it this way: Letting $F$ be as you specified, it follows that $F(0) geq 0$ while $F(1) le 0$. Furthermore, $F$ is also continuous. Therefore by the intermediate value theorem there is an $x in [0,1] = I$ s.t. $F(x) = 0$, or equvialently, $f(x) = x$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 16 at 20:40









          MikeMike

          4,621512




          4,621512












          • $begingroup$
            can you check mine now?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:45










          • $begingroup$
            Well, the writing really does need to be improved. And, the inequality $f(x) > x$ could hold for every point in $I setminus {1}$. So I would advise you to try again.
            $endgroup$
            – Mike
            Jan 16 at 20:50










          • $begingroup$
            The domain is compact!? So we cannot leave out the boundary point.
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:53










          • $begingroup$
            Does the result hold good in $(0,1)$? Can you help me out?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:56










          • $begingroup$
            To answer your question, the result does not necessarily hold in $(0,1)$. For example, consider the function $f(x)= 1/2 + x/2$. The idea for showing that it does hold in $[0,1]$ is this: The function $F$ is nonnegative at $x=0$ and it is nonpositive at $x=1$. so as $F$ is also continuous, somewhere in $[0,1]$ (even if it is at precisely one endpoint $x=0$ or the other endpoint $x=1$ or at both endpoints of $I$, it must take on the value 0.
            $endgroup$
            – Mike
            Jan 16 at 20:59




















          • $begingroup$
            can you check mine now?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:45










          • $begingroup$
            Well, the writing really does need to be improved. And, the inequality $f(x) > x$ could hold for every point in $I setminus {1}$. So I would advise you to try again.
            $endgroup$
            – Mike
            Jan 16 at 20:50










          • $begingroup$
            The domain is compact!? So we cannot leave out the boundary point.
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:53










          • $begingroup$
            Does the result hold good in $(0,1)$? Can you help me out?
            $endgroup$
            – Subhasis Biswas
            Jan 16 at 20:56










          • $begingroup$
            To answer your question, the result does not necessarily hold in $(0,1)$. For example, consider the function $f(x)= 1/2 + x/2$. The idea for showing that it does hold in $[0,1]$ is this: The function $F$ is nonnegative at $x=0$ and it is nonpositive at $x=1$. so as $F$ is also continuous, somewhere in $[0,1]$ (even if it is at precisely one endpoint $x=0$ or the other endpoint $x=1$ or at both endpoints of $I$, it must take on the value 0.
            $endgroup$
            – Mike
            Jan 16 at 20:59


















          $begingroup$
          can you check mine now?
          $endgroup$
          – Subhasis Biswas
          Jan 16 at 20:45




          $begingroup$
          can you check mine now?
          $endgroup$
          – Subhasis Biswas
          Jan 16 at 20:45












          $begingroup$
          Well, the writing really does need to be improved. And, the inequality $f(x) > x$ could hold for every point in $I setminus {1}$. So I would advise you to try again.
          $endgroup$
          – Mike
          Jan 16 at 20:50




          $begingroup$
          Well, the writing really does need to be improved. And, the inequality $f(x) > x$ could hold for every point in $I setminus {1}$. So I would advise you to try again.
          $endgroup$
          – Mike
          Jan 16 at 20:50












          $begingroup$
          The domain is compact!? So we cannot leave out the boundary point.
          $endgroup$
          – Subhasis Biswas
          Jan 16 at 20:53




          $begingroup$
          The domain is compact!? So we cannot leave out the boundary point.
          $endgroup$
          – Subhasis Biswas
          Jan 16 at 20:53












          $begingroup$
          Does the result hold good in $(0,1)$? Can you help me out?
          $endgroup$
          – Subhasis Biswas
          Jan 16 at 20:56




          $begingroup$
          Does the result hold good in $(0,1)$? Can you help me out?
          $endgroup$
          – Subhasis Biswas
          Jan 16 at 20:56












          $begingroup$
          To answer your question, the result does not necessarily hold in $(0,1)$. For example, consider the function $f(x)= 1/2 + x/2$. The idea for showing that it does hold in $[0,1]$ is this: The function $F$ is nonnegative at $x=0$ and it is nonpositive at $x=1$. so as $F$ is also continuous, somewhere in $[0,1]$ (even if it is at precisely one endpoint $x=0$ or the other endpoint $x=1$ or at both endpoints of $I$, it must take on the value 0.
          $endgroup$
          – Mike
          Jan 16 at 20:59






          $begingroup$
          To answer your question, the result does not necessarily hold in $(0,1)$. For example, consider the function $f(x)= 1/2 + x/2$. The idea for showing that it does hold in $[0,1]$ is this: The function $F$ is nonnegative at $x=0$ and it is nonpositive at $x=1$. so as $F$ is also continuous, somewhere in $[0,1]$ (even if it is at precisely one endpoint $x=0$ or the other endpoint $x=1$ or at both endpoints of $I$, it must take on the value 0.
          $endgroup$
          – Mike
          Jan 16 at 20:59




















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          Human spaceflight

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          Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e
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          Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

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          0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB]
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          File:DeusFollowingSea.jpg

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          File:DeusFollowingSea.jpg From Wikipedia, the free encyclopedia Jump to navigation Jump to search File File history File usage No higher resolution available. DeusFollowingSea.jpg ‎ (238 × 211 pixels, file size: 12 KB, MIME type: image/jpeg ) Summary [ edit ] Media data and Non-free use rationale Description Cover art of the Deus CD Following Sea Author or copyright owner Deus Source (WP:NFCC#4) rocksucker.co.uk Date of publication 2012 Use in article (WP:NFCC#7) Following Sea Purpose of use in article (WP:NFCC#8) to serve as the primary means of visual identification at the top of the article dedicated to the work in question. Not replaceable with free media because (WP:NFCC#1) n.a. Minimal use (WP:NFCC#3) Low res image used in infobox of the article on the album Respect for commercial opportunities (WP:NFCC#2) n.a. Fair use Fair use of co
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